$\lim_{x\to -1}\dfrac{5-\sqrt{3x+28}}{x+1}=$
Substituting $x=-1$ into $\dfrac{5-\sqrt{3x+28}}{x+1}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{5-\sqrt{3x+28}}{x+1} \\\\ &=\dfrac{5-\sqrt{3x+28}}{x+1}\cdot\dfrac{5+\sqrt{3x+28}}{5+\sqrt{3x+28}} \gray{\text{Rationalize the numerator}} \\\\ &=\dfrac{5^2-(3x+28)}{(x+1)(5+\sqrt{3x+28})} \\\\ &=\dfrac{-3\cancel{(x+1)}}{\cancel{(x+1)}(5+\sqrt{3x+28})} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{-3}{5+\sqrt{3x+28}} \text{, for }x\neq -1 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-1$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{5-\sqrt{3x+28}}{x+1}=\dfrac{-3}{5+\sqrt{3x+28}}$ for all $x$ -values in the interval $(-1.5,-0.5)$ except for $x=-1$. Therefore, $\lim_{x\to -1}\dfrac{5-\sqrt{3x+28}}{x+1}=\lim_{x\to -1}\dfrac{-3}{5+\sqrt{3x+28}}=-\dfrac{3}{10}$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -1}\dfrac{5-\sqrt{3x+28}}{x+1}=-\dfrac{3}{10}$.